1977 Canadian MO Problems/Problem 1
Revision as of 22:07, 25 July 2006 by Boy Soprano II (talk | contribs)
Problem
If prove that the equation
has no solutions in positive integers
and
Solution
Directly plugging and
into the function,
We now have a quadratic in
Applying the quadratic formula,
In order for both and
to be integers, the discriminant must be a perfect square. However, since
the quantity
cannot be a perfect square when
is an integer. Hence, when
is a positive integer,
cannot be.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.